Integrating using long division and completing the square
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Integration using long division
- [Voiceover] See if you can evaluate this integral right over here. Assuming you had a goal at it so let's work through this together. You probably realized that some of the traditional techniques that we've already had in our tool kits don't seem to be directly applicable, u-substitution and others. The key here to realize is we have a rational expression here where the numerator has the same degree or higher than the denominator. In this case, the numerator and the denominator have the same degree. Whenever you see something like that, it's probably a good idea to divide the denominator into the numerator. That's what this rational expression could be interpreted as. X minus five, divided by negative two x plus two. So it's a little bit of algebraic long division. To actually divide negative two x plus two into x minus five is to see if we can rewrite this in a way where we can evaluate the integral. So let's do that. We're going to take x minus five. X minus five, and divide negative two x plus two into that. Negative two x plus two. Look at the highest degree terms. How many times does negative two x go into x? Let's going to go negative 1/2 times. Negative 1/2 times two is negative one. Negative 1/2 times negative two x is this is going to be positive x. Just like that. Now we want to subtract this yellow expression from this blue expression, and so let's just, I'll just take the negative of this and then add. So I'm just going to take the negative of it and add. We are left with negative five plus one is negative four. You could say negative two x plus two goes into x minus five, negative 1/2 times with negative four left over. So we can rewrite this integral, our original integral as, we can rewrite it as negative 1/2 minus four over negative two x plus two, d x. Now let's see, looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two. All of these terms are divisible by two. Actually, we all have these negatives that always unnecessarily complicate things. Let's actually divide the numerator and the denominator by negative two. What are we going to have then? We divide the numerator by negative two. If this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's just going to become x. Then two divided by negative two is going to be minus one. So our original integral... This is just all algebra. Everything we've done so far is algebra. We just rewritten it using a little bit of algebraic long division. Our original integral has simplified to negative 1/2, and some might argue it's not simplified but it's actually much more useful for finding the integral. Negative 1/2 plus two over x minus one, d x. Now, how do we evaluate this? Well, the antiderivative of negative 1/2 is pretty straight forward. That's just going to be negative 1/2 x, plus the entire derivative of two over x minus one. You might reduce here ahead. The derivative of x minus one is just one so you could say that the derivative is sitting there. We can essentially use substitution in our heads and say, "Okay, let's just take the entire derivative, "we could say with respect to x minus one which will be "the natural log of the absolute value of x minus one." If all of that sounds really confusing, I'll let you do the u-substitution. If I were just trying to evaluate the integral two over x minus one, d x, I could see, okay, the derivative of x minus one is just one so I could say u is equal to x minus one, and then d u is going to be equal to d x. This is going to be, we can rewrite in terms of u as two, I'll just take the constant out two times the integral of one over u, d u, which we know as two times the natural log of the absolute value of u plus c. In this case, we know that u is x minus one. This is equal to two times the natural log of x minus one plus c. That's what we're going to have right over here. So plus two times the natural log of the absolute value of x minus one plus c. The plus c doesn't just come from this one, this general overtaking the integral of the whole thing. It could be some constant because if we go the other way, we take the derivative and the constant will go away. Let me just put the plus c right over there, and we are done.